【纽约区域赛】-E-A Rational Sequence （递归，满二叉树）

E   A Rational Sequence 
 

A sequence of positive rational numbers is defined as follows: 

 

An infinite full binary tree labeled by positive rational numbers is defined by: 

 

 The label of the root is 1/1. 

 The left child of label p/q is p/(p+q). 

 The right child of label p/q is (p+q)/q. 

 

The top of the tree is shown in the following figure: 



 

 

The sequence is defined by doing a level order (breadth first) traversal of the tree (indicated by the 

light dashed line). So that: 

 

F(1) = 1/1, F(2) = 1/2, F(3) = 2/1, F(4) = 1/3, F(5) = 3/2, F(6) = 2/3, … 

 

Write a program which finds the value of n for which F(n) is p/q for inputs p and q. 

Greater New York Regional E • A Rational Sequence

 

Input 

 

The first line of input contains a single integer P, (1  P  1000), which is the number of data sets 

that follow. Each data set should be processed identically and independently. 

 

Each data set consists of a single line of input. It contains the data set number, K, a single space, the 

numerator, p, a forward slash (/) and the denominator, q, of the desired fraction. 

 

 

Output 

 

For each data set there is a single line of output. It contains the data set number, K, followed by a 

single space which is then followed by the value of n for which F(n) is p/q. Inputs will be chosen so n 

will fit in a 32-bit integer. 

 

 

 

Sample Input Sample Output 

4 

1 1/1 

2 1/3 

3 5/2 

4 2178309/1346269 

 

1 1 

2 4 

3 11 

4 1431655765 

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
LL h;
LL search(LL a,LL b)		//返回当前节点，前面节点的个数 (再加上上面所有节点的个数((1<<h)-1)即可)
{
h++;
if(a==b)			//当返回根结点时，根节点前面没有节点，所以是 0
return 0;
if(a>b)
return (search(a-b,b))*2;
if(a<b)
return (search(a,b-a))*2-1;
}
int main()
{
int u;
LL k,a,b;
scanf("%d",&u);
while(u--)
{
h=0;
scanf("%lld %lld/%lld",&k,&a,&b);
printf("%lld %lld\n",k,search(a,b)+((LL)1<<h)-1);
}
return 0;
}